Problem: Given a string of length n s = s1, s2,…, sn and m queries. Each query qk (1 ≤ k ≤ m) is one of four types, "L ++", "L-", "R ++", "R-", and l [for the kth query qk. k] and r [k] are defined below.

* L ++: l [k] = l [k-1] + 1, r [k] = r [k-1]

* L-: l [k] = l [k-1] -1, r [k] = r [k-1]

* R ++: l [k] = l [k-1], r [k] = r [k-1] +1

* R-: l [k] = l [k-1], r [k] = r [k-1] -1




However, l [0] = r [0] = 1.

At this time, how many kinds of character strings are created for m substrings sl [k], sl [k] + 1,…, sr [k] -1, sr [k] (1 ≤ k ≤ m). Answer if you can.

Constraints

* The string s consists of a lowercase alphabet

* 1 ≤ n ≤ 3 x 105

* 1 ≤ m ≤ 3 × 105

* qk (1 ≤ k ≤ m) is one of "L ++", "L-", "R ++", "R-"

* 1 ≤ l [k] ≤ r [k] ≤ n (1 ≤ k ≤ m)

Input

Input is given in the following format

> n m
> s
> q1
> q2
>…
> qm
>

Output

Output the solution of the problem on one line

Examples

Input

5 4
abcde
R++
R++
L++
L--


Output

3


Input

4 6
abab
R++
L++
R++
L++
R++
L++


Output

4


Input

10 13
aacacbabac
R++
R++
L++
R++
R++
L++
L++
R++
R++
L--
L--
R--
R--


Output

11

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
//const ll mod = 1000000007;
template<class Z> Z rng(Z a, Z b) {
  static mt19937 mt(chrono::steady_clock::now().time_since_epoch().count());
  return uniform_int_distribution<Z>(a, b - 1)(mt);
}

struct RollingHash {
  static constexpr uint64_t P0 = 4e9 + 7;
  static constexpr uint64_t P1 = 4e9 + 9;
  static uint64_t B0;
  static uint64_t B1;
  static vector<uint64_t> powB0;
  static vector<uint64_t> powB1;
  const int n;
  vector<uint64_t> h0;
  vector<uint64_t> h1;
  template<class Itr> RollingHash(Itr first, Itr last) : n(distance(first, last)), h0(n + 1), h1(n + 1) {
    for (int i = 0; i < n; ++i, ++first) {
      h0[i + 1] = (h0[i] * B0 + *first) % P0;
      h1[i + 1] = (h1[i] * B1 + *first) % P1;
    }
    while (powB0.size() <= n) {
      powB0.push_back(powB0.back() * B0 % P0);
      powB1.push_back(powB1.back() * B1 % P1);
    }
  }
  uint64_t get0(int l, int r) { return (h0[r] + (P0 - h0[l]) * powB0[r - l]) % P0; }
  uint64_t get1(int l, int r) { return (h1[r] + (P1 - h1[l]) * powB1[r - l]) % P1; }
  bool eq(int l0, int r0, int l1, int r1) {
    return get0(l0, r0) == get0(l1, r1) and get1(l0, r0) == get1(l1, r1);
  }
  template<class Itr> static uint64_t get0(Itr first, Itr last) {
    uint64_t res = 0;
    while (first != last) {
      res = (res * B0 + *first++) % P0;
    }
    return res;
  }
  template<class Itr> static uint64_t get1(Itr first, Itr last) {
    uint64_t res = 0;
    while (first != last) {
      res = (res * B1 + *first++) % P1;
    }
    return res;
  }
};
uint64_t RollingHash::B0 = rng<unsigned>(1, RollingHash::P0);
uint64_t RollingHash::B1 = rng<unsigned>(1, RollingHash::P1);
vector<uint64_t> RollingHash::powB0{1};
vector<uint64_t> RollingHash::powB1{1};


int main() {
    //cout.precision(10);
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N, M;
    cin >> N >> M;
    string S;
    cin >> S;
    RollingHash RHS(S.begin(), S.end());
    set<l_l> st;
    int l = 0;
    int r = 1;
    while(M--) {
        string q;
        cin >> q;
        if(q == "R++") r++;
        if(q == "R--") r--;
        if(q == "L++") l++;
        if(q == "L--") l--;
        ll one = RHS.get0(l, r);
        ll two = RHS.get1(l, r);
        st.insert({one, two});
    }
    cout << st.size() << endl;
    return 0;
}